For the same objective, find the ratio of the least separation between two points to be distinguished by a microscope for light of 5000 Å and electrons accelerated through 100 V used as the illuminating substance.

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Vishal Baghel | Contributor-Level 10

7 months ago

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- resolving power =1/d = $\frac{2 s i n β}{1.22 λ}$ = d_min= $\frac{1.22 λ}{2 s i n β}$ where $λ$ is the wavelength of light

So d_min= $\frac{1.22 * 5500 * 10^{- 10}}{2 s i n β}$

$λ$ = $\frac{12.27}{\sqrt[]{V}}$ = 0.12 $*$ 10^-9m

$\frac{d' m i n}{d m i n}$ = $\frac{0.12 * 10^{- 9}}{5500 * 10^{- 10}}$ = 0.2 $*$ 10^-3

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Physics Ncert Solutions Class 12th 2023

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